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conference - sets

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Manager
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conference - sets [#permalink] New post 14 May 2011, 21:49
00:00
A
B
C
D
E

Difficulty:

  5% (low)

Question Stats:

100% (02:06) correct 0% (00:00) wrong based on 6 sessions
At a certain conference, 72% of the attendees registered at least
two weeks in advance and paid their conference fee in full. If
10% of the attendees who paid their conference fee in full did
not register at least two weeks in advance, what percent of
conference attendees registered at least two weeks in
advance?
(A) 18.0%
(B) 62.0%
(C) 79.2%
(D) 80.0%
(E) 82.0%
[Reveal] Spoiler: OA
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Re: conference - sets [#permalink] New post 14 May 2011, 23:52
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attendees = 100
people who paid the fees in full = x

0.1x = who paid the fee in full but did not do 2 weeks in advance.
0.9x = people who paid the fee in full and paid atleast 2 weeks in advance = 72

x= 72/0.9 = 80

thus people who paid fee and not 2 weeks in advance = 0.1*80 = 8
so

Fee !Fee
2 weeks advance 72 20 92
! 2 week advance 8 0 8

total = 100

so 92%
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Re: conference - sets [#permalink] New post 15 May 2011, 00:09
fanatico wrote:
At a certain conference, 72% of the attendees registered at least
two weeks in advance and paid their conference fee in full. If
10% of the attendees who paid their conference fee in full did
not register at least two weeks in advance, what percent of
conference attendees registered at least two weeks in
advance?
(A) 18.0%
(B) 62.0%
(C) 79.2%
(D) 80.0%
(E) 82.0%


Looks like this question is not complete. The required percent cannot be determined with given data. The answer can range from 72% to 92%. Could you please confirm?
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Re: conference - sets [#permalink] New post 15 May 2011, 01:51
My take :

0.72A -> At least 2 weeks in advance and paid conference fee in full.

FA -> Paid the conference fee in full

0.10FA -> Not 2 weeks in advance paid but conference fee in full

=> 0.90FA -> 2 Weeks in advance and paid the conference fee in full

0.90FA = 0.72A

FA/A = 72/90 * 100

= 80%

Answer - D
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Manager
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Re: conference - sets [#permalink] New post 15 May 2011, 07:16
I got the same answer as amit2k9.

@subhashgosh
FA/A is not wat is asked, I think.
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Re: conference - sets [#permalink] New post 17 May 2011, 10:39
fluke wrote:
fanatico wrote:
At a certain conference, 72% of the attendees registered at least
two weeks in advance and paid their conference fee in full. If
10% of the attendees who paid their conference fee in full did
not register at least two weeks in advance, what percent of
conference attendees registered at least two weeks in
advance?
(A) 18.0%
(B) 62.0%
(C) 79.2%
(D) 80.0%
(E) 82.0%


Looks like this question is not complete. The required percent cannot be determined with given data. The answer can range from 72% to 92%. Could you please confirm?


Exactly
we dont have any bifurcation of people who did not pay the full fee (how many registered in 2 week advance and how many did not)

without knowing that we can't reach the desired answer that how many registered in 2 weeks advance
it could be 72% +0% (minimum) to 72% + 20% (maximum)

please clarify :? :?
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Re: conference - sets [#permalink] New post 17 May 2011, 16:09
Something wrong with the question. we need additional data to solve the problem.
Director
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Re: conference - sets [#permalink] New post 17 May 2011, 16:12
@amit2k9,

how can you say that people who didnt register two weeks in advance and not paid in full =0 ? Did you assume that?



amit2k9 wrote:
attendees = 100
people who paid the fees in full = x

0.1x = who paid the fee in full but did not do 2 weeks in advance.
0.9x = people who paid the fee in full and paid atleast 2 weeks in advance = 72

x= 72/0.9 = 80

thus people who paid fee and not 2 weeks in advance = 0.1*80 = 8
so

Fee !Fee
2 weeks advance 72 20 92
! 2 week advance 8 0 8

total = 100

so 92%
Senior Manager
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Re: conference - sets [#permalink] New post 17 May 2011, 16:27
yes between 72 and 92%, so answer could be any of C, D and E.
Re: conference - sets   [#permalink] 17 May 2011, 16:27
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